How to proof $ \underset{n\to\infty}\lim \int_0^1 f(t)\, \mathrm{sgn}\big(\sin (2\pi n t)\big)\,dt = 0$?

Question

I was wondering if you could provide hints which could lead me to a rigorous proof for the following:

Given $\,f\in L^1([0,1])$, then

$$ \underset{n\to\infty}\lim \int_0^1 f(t)\, \mathrm{sgn}\big(\sin (2\pi n t)\big)\,dt = 0. $$

Thank you.

Answer 1

First prove it for $f \in C[0,1]$, then as these functions are dense in $L^1[0,1]$ you are done.

Notice if you denote the quantity you're taking the limit of by $I_n$, for $f\in C[0,1]$, and you extend $f$ periodically so $f(y)$ makes sense for all $y \in \mathbb R$ (don't forget this extended function may not be continuous at integer points of $\mathbb R$, but it turns out not to matter), $$\begin{align} I_n &= \int_0^1 f(t)\, \mathrm{sgn}(\sin (2\pi nt ))\, dt \\ &= \int_{1/2n}^{1+1/2n}f(t-1/2n)\,\mathrm{sgn}(\sin (2 \pi nt -\pi))\, dt &\text{ (Substitute } t \text{ with } t-1/2n \text{)}\\ &= -\int_0^1 f(t-1/2n)\,\mathrm{sgn}(\sin(2\pi t))\, dt &\text{(Periodicity and property of } \sin \text{)}, \end{align}$$ Hence $$I_n= \frac{1}{2} \int_0^1 \left[f(t) - f(t-1/2n)\right]\mathrm{sgn}(\sin(2\pi nt))\, dt.$$ Now think why this integral converges to zero.

Answer 2

Hints.

a. If $f\in L^1[0,1]$, then $f$ can be approximated, in the $L^1-$norm by step functions, i.e., functions for which there are $1=s_0<s_1<s_{n-1}<s_n=1$, such that $f$ is constant in every open subinterval.

b. Due to it suffices to show the question for $f=\chi_{[a,b]}$, where $0\le a<b\le 1$.

c. Finally observed that $$ \int_0^1 \mathrm{sgn}\big(\sin(2\pi nx)\big)\chi_{[a,b]}(t)\,dt =\int_a^b \mathrm{sgn}\big(\sin(2\pi nx)\big)\,dt, $$ the this last interval is absolutely less that $2/n$.